#include <vector>
#include <unordered_map>
#include <iostream>
using namespace std;
class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        unordered_map<int, int> umap; //key:a+b的数值，value:a+b数值出现的次数
        // 遍历大A和大B数组，统计两个数组元素之和，和出现的次数，放到map中
        //注意：umap初始值为空，所以umap[a+b]++是1，如果再次出现相同的a+b，那么umap[a+b]++就是2
        for (int a : A) {
            for (int b : B) {
                umap[a + b]++;
                cout <<"a="<<a<<"b="<<b<<"umap[" <<a+b<<"] = "<<umap[a+b]<<endl;
            }
        }
        //遍历umap
        for(auto i:umap)
        {
            cout<<"i.first="<<i.first<<"   i.second="<<i.second<<endl;
        }
        int count = 0; // 统计a+b+c+d = 0 出现的次数
        // 再遍历大C和大D数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来。
        for (int c : C) {
            for (int d : D) {
                //如果在umap中找到0-(c+d)这个key，那么就把对应的value也就是出现次数统计出来
                if (umap.find(0 - (c + d)) != umap.end()) {
                    count += umap[0 - (c + d)];
                }
            }
        }
        return count;
    }
};
// 单元测试
int main() {
    Solution solution;
    vector<int> A = {1, 2};
    vector<int> B = {-2, -1};
    vector<int> C = {-1, 2};
    vector<int> D = {0, 2};
    
    int result = solution.fourSumCount(A, B, C, D);
    cout << "The count of four sum is: " << result << endl; // 应输出 2
    
    return 0;
}